harmonic motion p221, november 22 nd, 2013. review of simple harmonic motion system at rest displace...

Harmonic Motion

P221, November 22nd, 2013

Review of Simple Harmonic Motion

• System at rest

• Displace mass,stretches spring

• Restoring force is proportionalto displacement

0 x

F

More Review

• No external forces energy conserved

• Kinetic is converted to potential, vice versa

• Velocity at ends is 0“turning point”

• Fastest at center• Frequency is constant

t

F v

Sines & Cosines

• Restoring force is linear AND in opposite direction to displacement

Sines & Cosines


• Combination of sines & cosines can solve this

Sines & Cosines



• Angular frequency is ALWAYS (and independent of amplitude)

• “Coordinate” could be x or q or anything else

Sines & Cosines



• Angular frequency is ALWAYS (and independent of amplitude)

• “Coordinate” could be x or q or anything else

Has a physical interpretation

Simple Pendulum (q coordinate)

Rotational Oscillation (q)

• Torque is proportional to angular displacement

1-D Spring and Block (x)

• Force is proportional to positional displacement

q I wire

k could be mg or torsional

strength


Rotational Oscillation (q) 1-D Spring and Block (x)

q I wire


Rotational Oscillation (q)

• Torque is proportional to angular displacement

1-D Spring and Block (x)

• Force is proportional to positional displacement

q I wire

Clicker Question

• Systems 1 & 2 are oscillating at their own frequencies. We then double the masses. Do the frequencies change?

A) Both changeB) Neither changeC) Only system 1 changesD) Only system 2 changes

q

System 1

System 2

m

m

Clicker Question: Discussion

• System 1: k is a spring constant that is independent of mass

q

System 1

System 2

m

m


• System 1: k is a spring constant that is independent of mass

• System 2: both restorative force and moment of inertia are proportional to mass

q

System 1

System 2

m

m

Physical Pendulum

q RC

M

Mg

XCM

Physical Pendulum

q RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

XCM

Physical Pendulum

Iq RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

XCM

Physical Pendulum

Iq

CMMgXXCM

RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

Physical Pendulum

Forsmall q

Iq

CMMgXXCM

RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

CMMgR

Physical Pendulum

Forsmall q

I

2

2

dt

dI

q

CMMgXXCM

RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

CMMgR

Physical Pendulum

Forsmall q

I

2

2

dt

dI

q

CMMgXXCM

RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

CMMgR

I

MgR

dt

d CM2

2

Physical Pendulum

Forsmall q

I

2

2

dt

dI

q

CMMgXXCM

RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

CMMgR

I

MgR

dt

d CM2

2

22

2

dt

d

Physical Pendulum

Forsmall q

I

2

2

dt

dI

q

CMMgXXCM

RC

M

Mg

q

arc-length = RCM q

XCM

RC

M

I

MgRCM

CMMgR

I

MgR

dt

d CM2

2

22

2

dt

d

The Simple Pendulum IS a Physical Pendulum

The general case

CM

pivot

q

RCM

I

MgRCM

The simple case

qL

2ML

MgL

L

g

A Specific Case: Stick Pendulum

M

pivot

q

RCM

CM


M

pivot

q

RCM

CM

I

MgRCM


2

L

2

3

1ML

M

pivot

q

RCM

CM

I

MgRCM


L

g

32

2

L

2

3

1ML

M

pivot

q

RCM

CM

I

MgRCM


L

g

32

2

L

2

3

1ML

M

pivot

q

RCM

CM

I

MgRCM

Same period

L3

2

L

Clicker Question

• In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attachedto the center of the samestick. Which pendulum hasthe longer period?

A) Case 1B) Case 2C) Same

Case 1

m

Case 2

m

m

Case 2

1

2L

Clicker Question: Prelude

• In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attached to a string of length L/2?

• Which as the longer period?

A) Case 1B) Case 2C) Same

L

Case 1

m

L

Case 1

L

g

32

m

Case 2

1

2L L

g

21

Prelude Answer

• Remember period is inversely proportional to rotational frequency w

therefore

Clicker Question: Prelude 2

• We know that T1 > T2. Now suppose these pendula are “glued” together from the same pivot. What is the new period?

A) T1B) T2C) In Between

mT1

T2

T1 > T2

mm

+ =


• We know that T1 > T2 and T of the “glued” pendulum is in between. We have proven T1 is the longest. But, let’s calculate in detail!

mT1

T1 > T2

T2

mm

Case 2m

m

mCase 1

I

MgRCM

Clicker: Detailed Answer

2

Lmg

Case 2m

m

mCase 1

I

MgRCM


22

Lmg

2

Lmg

Case 2m

m

mCase 1

I

MgRCM


22

Lmg

2

3

1mL 2

22

12

7

23

1mL

LmmL

Case 2m

m

mCase 1


L

g

32

L

g

127

Mechanics Lecture 21, Slide 43

y yk

mg


y yk

mg

ykmg y

mgk


y yk

mg

ykmg y

mgk

m

k

At t = 0, y = 0, moving down


tAty sin)(

tAtv cos)( tAta sin)( 2


Use energy conservation to find A


tAty sin)(


22max 2

1

2

1kAmv


Use energy conservation to find A


tAty sin)(


22max 2

1

2

1kAmv

k

mvA max

tAtv cos)(


tAta sin)( 2

Aa 2max

Or similarly

m

kA

m

yk

m

Fa

maxmax

max


)(sin)()( tkAtkytF


tAty sin)(

2

2

1kyU

harmonic motion p221, november 22 nd, 2013. review of simple harmonic motion system at rest displace...

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